Question

In a $△ABC$, let $\angle C=\frac{\pi }{2}$.

If $r$ is the inradius and $R$ is the circumradius of the triangle, then $2(r+R)$ is equal to ?

• A. $a+b$
• B. $b+c$
• C. $c+a$
• D. $a+b+c$

Answer 1

The correct option is A $a+b$

The hypotenuse of the right triangle $ABC$ is $AB$.

We take $C$ at the origin and $CB$ along x-axis and $CA$ along y-axis.

The mid-point $M(\frac{a}{2},\frac{b}{2})$ of $AB$ is the circumcenter of the triangle.

Therefore, $R^2={MC}^2=\frac{1}{4}(a^2+b^2)=\frac{1}{4}{c}^2\Rightarrow R=\frac{c}{2}$

Next, $r=(s-c)\tan \frac{C}{2}=(s-c)\tan \frac{\pi }{4}=s-c$

Thus, $2(r+R)=2r+2R=2s-2c+c=a+b+b-2c+c=a+b$